Asked by Matt

A basketball was kicked 20.0 m above horizontal with an initial velocity of 120.0 m/s at an angle of 35.0˚ above the horizontal.
(a) How long is the rocket in air before falling?
(b) What is the horizontal range for the rocket?
(c) With what speed does the rocket hit the ground?

Answers

Answered by Henry
Vo = 120m/s[35o].
Xo = 120*Cos35 = 98.3 m/s.
Yo = 120*sin35 = 68.8 m/s.

a. Y = Yo + g*Tr = 0.
68.8 - 9.8Tr = 0, Tr = 7.02 s. = Rise time.

h = ho + Yo*Tr + 0.5g*Tr^2 =
20 + 68.8*7.0 - 4.9*7.0^2 = 261.5 m. above gnd.

0.5g*Tf^2 = 261.5.
4.9*Tf^2 = 261.5, Tf = 7.31 s. = Fall time.

Tr+Tf = 7.02 + 7.31 = 14.33 s. = Time in air.

b. Range = Vo^2*sin(2A)/g = 120^2*sin(70)/9.8 = 1381 m.

c. Y = Yo + g*Tf = 0 + 9.8*7.31 = 71.6 m/s. = Ver. component.

V = Sqrt(Xo^2+Y^2) = Sqrt(98.3^2+71.6^2) =
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